$\displaystyle

\int (xsinx^2cosx^2),dx

$

all i need are steps leading to the answer

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- Feb 25th 2008, 01:45 PMjliddellintegration part 2
$\displaystyle

\int (xsinx^2cosx^2),dx

$

all i need are steps leading to the answer - Feb 25th 2008, 01:46 PMKrizalid
Start by makin' $\displaystyle u=x^2.$

- Feb 25th 2008, 01:58 PMjliddellintegration part 2
ok not quite getting it. can you help me out a little more please.

- Feb 25th 2008, 02:00 PMKrizalid
$\displaystyle u=x^2\implies du=2x\,dx,$

$\displaystyle \int {x\sin x^2 \cos x^2 \,dx} = \frac{1}

{2}\int {\sin u\cos u\,du} = \frac{1}

{4}\int {\sin 2u\,du} .$ - Feb 26th 2008, 10:57 AMsimsima_1needing a little more help w/ this....
ok so the answer is

$\displaystyle

(x)tan^{-1}(x)-\frac{1}{2}ln({x^2}+1)+C

$

and i'm not sure how i would get that using

$\displaystyle

{1/4} \int sin2u

$ - Feb 26th 2008, 11:20 AMJhevon